25N Sideways Force on 10kg Box: F = m*a Formula Error

AI Thread Summary
A 25N sideways force is applied to a 10kg box moving at a constant speed of 5 m/s. The discussion clarifies that if the box is moving with constant velocity, the net force acting on it is zero, indicating no acceleration. The presence of friction was questioned but not confirmed, suggesting it may not be a factor. The hint provided encourages careful consideration before applying equations. Ultimately, the conclusion is that the net force is zero due to the absence of acceleration.
mystry4
Messages
14
Reaction score
0
If you have a 25N sideways force on a 10kg box that moves at speed of 5 m/s, what is the formula? I used F = m*a and got it wrong. :confused:
 
Physics news on Phys.org
What are you being asked to find out?
 
I am being asked to find the net force on the box.
 
well, so far the only force acting on it is the 25 N,
Is there friction?
 
Also, if the box is moving with constant velocity, i.e. no acceleration, then the net force is zero.
 
Well that makes sense...and probably why a HINT was put in there (HINT: Think about this a bit before writing down any equations!)
So, I amanswering that the net force is O because of continual velocity and no acceleration.
Thank you for your help
 
Thread 'Collision of a bullet on a rod-string system: query'

Thread 'Collision of a bullet on a rod-string system: query'

In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Similar threads

A steel shuttle is pulled along a rail, what is its initial acceleration?
Replies
3
Views
2K
What will be the acceleration of a 10kg block w.r.t. the ground frame?
Replies
6
Views
2K
Simple Work Problem Disagreeing with Kinematics
Replies
4
Views
1K
Acceleration & Normal Force of a Box: 4 m/s2 & 418 N
Replies
9
Views
2K
Static Friction Between a Box and the Floor
Replies
11
Views
3K
Motion on a rough inclined plane
Replies
5
Views
2K
How Does Impulse Relate to Momentum Change in Physics Problems?
Replies
7
Views
2K
How do you calculate the margin of error in Physics?
Replies
10
Views
786
Using Newton's Laws: Friction, Circular Motion, Drag Forces
Replies
1
Views
1K
Calculate the Acceleration of this Car
Replies
7
Views
921

Hot Threads

Recent Insights

Back
Top